Q. 73.7( 13 Votes )

# In Fig. 6.14, DE

It is given to us –

DE || QR and n is the transversal intersecting DE and QR at points A and B respectively.

We have to find APB.

AP and BP are the bisectors of EAB and ABR respectively.

EAP = PAB

EAB = 2 × PAB - - - - (i)

Also, RBP = PBA

RBA = 2 × PBA - - - - (ii)

Since, DE || QR,

RBA = EAn (Corresponding angles)

RBA = 180° - EAB (By linear pair axiom, EAn + EAB = 180°)

Using equations (i) and (ii) in the above equation,

2 × PBA = 180° - (2 × PAB)

PBA = 90° - PAB (Dividing both sides by 2)

PBA + PAB = 90° - - - - (iii)

Now, in ΔPAB,

PBA + PAB + APB = 180° (Sum of three angles of a triangle is 180°)

90° + APB = 180° [From equation (iii)]

APB = 90°

Thus, the value of APB is 90°.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Introduction to Lines and Angles60 mins
Bonus Questions - 156 mins
Bonus Questions - 257 mins
Revision on Angle sum Property of Triangles44 mins
Sphere and Hemisphere42 mins
Speed and Velocity42 mins
Arc of Circle And Related IMP Qs40 mins
Surface Area and Volume of solids Revision43 mins
IMP Theorems And Their Application43 mins
Circles- All kinds of Questions36 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

In the given figuRS Aggarwal & V Aggarwal - Mathematics

In Fig 8.117, RD Sharma - Mathematics

In Fig 8.116, if RD Sharma - Mathematics

In Fig. 8.136, ifRD Sharma - Mathematics

In Fig. 8.142, ifRD Sharma - Mathematics

In Fig. 8.138, ifRD Sharma - Mathematics

In Fig. 8.137, ifRD Sharma - Mathematics

In Fig. 8.131, ifRD Sharma - Mathematics

In Fig 8.121, armRD Sharma - Mathematics

In Fig. 8.139, ifRD Sharma - Mathematics